Question: Let $f(x) = 3x^{2}+6x-2$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $3x^{2}+6x-2 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 3, b = 6, c = -2$ $ x = \dfrac{-6 \pm \sqrt{6^{2} - 4 \cdot 3 \cdot -2}}{2 \cdot 3}$ $ x = \dfrac{-6 \pm \sqrt{60}}{6}$ $ x = \dfrac{-6 \pm 2\sqrt{15}}{6}$ $x =\dfrac{-3 \pm \sqrt{15}}{3}$